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Let $R$ be an Integral Domain satisfying Ascending Chain Condition for Principal Ideals. If $I$ is a proper ideal of $R$, does $R/I$ have ACCP?

This statement is False but I do not have a counter example to show for it.

A suggested solution (from classmate) was to consider $R = \mathbb Z[x_1,x_2,x_3,...]$ and let $I=(x_1-x_2^2,x_2-x_3^2,x_3-x_4^2,...)$. However I can't seem to see how this is a counterexample.

Could someone explain why $R$ has ACCP in the first place and why $R/I$ does not have ACCP.

Jhon Doe
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  • If $A$ is a ufd, then so is $A[x]$. And we can look at polynomial rings in several variables as polynomials in one variable with coefficients in $A[x_1,...,x_n]$. So if $A$ is a ufd, so is $A[x_1,...,x_n]$. Then, since it is a ufd, it has the ascending chain condition on p.i.s. – take008 Mar 23 '18 at 06:56
  • But this has infinite indeterminates – Jhon Doe Mar 23 '18 at 07:43
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    Oh I see. Then you can use this: https://math.stackexchange.com/questions/1266784/is-a-polynomial-ring-over-a-ufd-in-countably-many-variables-a-ufd – take008 Mar 23 '18 at 08:07
  • Thanks. Any idea for the second part? – Jhon Doe Mar 23 '18 at 08:09
  • No, sorry. I’m not good at finding counter examples. I’ll give it a shot tomorrow after I’m done with finals. – take008 Mar 23 '18 at 08:14
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    The point of the construction of $R/I$ is that it contains a square root chain: ${x_{k+1}}^2=x_k$ for all $k=1,2,\dots$ – Berci Mar 23 '18 at 08:28
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    For the second part: the chain of ascending principal ideals $(x_1)\subset(x_2)\subset(x_3)\subset\cdots$ is strict. – user26857 Mar 23 '18 at 15:27

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